If you look closer at the docs, will see that the success
callback is passed 4 arguments
1.) responseText or responseXML value (depending on the value of the dataType option).
2.) statusText
3.) xhr (or the jQuery-wrapped form element if using jQuery < 1.4)
4.) jQuery-wrapped form element (or undefined if using jQuery < 1.4)
Use the last argument instead of $(this)
for current form instance
Update by asker : working code from above answer I got is following
Instead of $(this)
I could do following
$('form').append('<div class="feedback" />');
var options = { success: showResponse };
$('form').ajaxForm(options);
function showResponse(responseText, statusText, xhr, $form)
{
if(statusText == 'success')
$form.children('.feedback').html(responseText);
}